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(C) The center of mass $(COM)$ of the projectile would have landed at the range $R = 100 \ m$. Since the explosion is an internal force,the $COM$ cont

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$250$

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(C) The center of mass $(COM)$ of the projectile would have landed at the range $R = 100 \ m$. Since the explosion is an internal force,the $COM$ continues to follow the original parabolic path. Let the masses of the three fragments be $m_1 = m_2 = m_3 = m$. Let their positions at the time of landing be $x_1, x_2, x_3$. Given: $x_1 = 0$ (returns to projection point),$x_2 = R/2 = 50 \ m$ (comes to rest at the highest point,so it falls vertically down),and $x_{COM} = R = 100 \ m$. The position of the center of mass is given by $x_{COM} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}$. Substituting the values: $100 = \frac{m(0) + m(50) + m(x_3)}{3m}$. $300 = 50 + x_3$. $x_3 = 300 - 50 = 250 \ m$.

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